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zheka24 [161]
3 years ago
11

What are real images? Are the upright or inverted?

Physics
1 answer:
Mars2501 [29]3 years ago
6 0

Answer:

A real image can be viewed when a screen is placed in the plane of convergence. Real images are inverted... HOPE THIS HELPS!!!

Explanation:

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An airplane accelerates down a runmway at 2.9 m/s/s for 38.5 seconds
Serhud [2]

Given,

the initial velocity = 0 m /s.

acceleration = 3.20 m / s^2

time = 32.8 s

According to laws of motion.

s = ut + 1/2 at ^2

s = 1/2 at²

s=1/2(3.20)(32.8)²

s= 1721.344 m

the distance traveled before takeoff is 1731.3m

8 0
3 years ago
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An ice cube is placed on a hot stove. Which of these statements best describes how heat moves between the ice cube and the stove
Tanzania [10]
B! Conduction is touch, so the heat traveled through touch from th stove to the ice cube, therefore melting it
6 0
3 years ago
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The built in flash in a compact camera is usally capable of giving correct exsposure for distance up to how many meters?
Brut [27]

Answer:

An on-camera flash is an indispensible accessory for many photographers; it provides additional light when conditions become too dark to handhold your camera comfortably, allows you to achieve more balanced exposures in daylight conditions, permits freezing of fast-moving subjects and can also be used to control or trigger other flash light sources. Additionally, a flash can be used as a highly effective creative tool to establish an aesthetic that elevates your imagery when lighting conditions are considered less than stellar. The benefits of an external on-camera flash far outweigh those provided by a built-in camera flash, while the only drawback is keeping an additional piece of equipment.

On-Camera Flash versus Off-Camera Flash versus In-Camera Flash

The term on-camera flash simply refers to a type of strobe light (flash) that can connect directly with your camera. While it is referred to as “on-camera” this does not require the flash to be physically mounted on your camera. On-camera flashes can, and often are, used off-camera. This differs from other strobe-light sources, such as studio pack strobes and monolights in that these types of strobes are not meant to be physically connected to your camera (except under rare and unusual circumstances involving convoluted methods of adaptation). Additionally, on-camera flashes usually have a self-contained power supply, although external power sources can sometimes be used to improve performance or battery life.

On-camera external flash also refers to the type of external flash that can be used on your camera, compared to a built-in flash that is integrated into many cameras. An on-camera external flash performs better than a built-in flash in almost every regard with the one exception that it is not built into your camera. The ability to take the flash off your camera results in a significantly greater number of lighting options; far more than simply providing a blast of flat light to the scene to facilitate an adequate exposure. It is often not desirable to have your flash pointed squarely at the scene at hand; more often than not you will want to bounce the flash light off other surfaces and point in other directions to control the look of your flash. When using an in-camera flash, you are forced to use the flash at the given angle from which it extends.

Most built-in flashes are also located near the camera lens, which can often result in the red-eye effect  when photographing subjects in dimly lit conditions. Red-eye occurs because pupils dilate in dim light, the built-in flash is aligned with the lens's optical axis, its beam enters the eye and reflects back at the camera from the retina at the rear of the eye, which is quite red. Being able to use an on-camera flash source off-camera, from a different angle, will help to eliminate the red-eye effect in your photographs of people.

Guide Numbers, Manual Usage, Controlling Flash Power and Sync Speeds

Before delving into the automatic technology that is contained within most contemporary flashes, it is best to understand how to manually control and grasp a flash’s power. This is directly related to having an understanding of exposure ratios—how shutter speeds and apertures affect and balance each other—even though auto-exposure metering is available and often utilized for determining the best exposure settings.

Explanation:

8 0
3 years ago
The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 77.0 nC. The plates are in va
WINSTONCH [101]

Answer:

Part A: 7500 V

Part B: 2.899×10⁻³ m²

Part C: 10.27 pF or 10.27×10⁻¹² F

Explanation:

Part A:

Applying,

E = V/d................ Equation 1

Where E = electric field intensity between the plates, V = potential difference between the plates, d = distance of separation between the plates

make V the subject of the equation above,

V = Ed............. Equation 2

Given: E = 3.0×10⁶ V/m, d = 2.5 mm = 2.5×10⁻³ m

Substitute into equation 2

V =  3.0×10⁶ (2.5×10⁻³ )

V = 7.5×10³ V

V = 7500 V

Part B:

Using,

E = Q/(e₀A).................... Equation 3

Where Q = Charge on each plate of the capacitor, A = Area of each plate, e₀ = constant = dielectric = permitivity of free space

make A the subject of the equation,

A = Q/(e₀E).............. Equation 4

Given: Q = 77 nC = 77×10⁻⁹ C, E = 3.0×10⁶ V/m

Constant: e₀ = 8.854×10⁻¹² F/m

Substitute into equation 4

A = 77×10⁻⁹/(8.854×10⁻¹²× 3.0×10⁶)

A = 77×10⁻⁹/(26.562×10⁻⁶)

A = 2.899×10⁻³ m²

A = 2.899×10⁻³ m².

Part C:

Using,

Q = CV.................. Equation 5

Where C = Capacitance of the capacitor

make C the subject of the equation

C = Q/V.............. Equation 6

Given: Q = 77 nC = 77×10⁻⁹ C, V = 7500 V

Substitute into equation 6

C = 77×10⁻⁹/7500

C = 10.27×10⁻¹² F

C = 10.27 pF

5 0
3 years ago
A bridge to be fabricated of steel girders is designed to be 500 m long and 12 m wide at ambient temperature (assumed 20°C). Exp
andrew-mc [135]

Answer:

23

21.7391304348 m

Explanation:

L = Initial length = 500 m

\Delta T = Change in temperature = 40-(-35)

\alpha = Coefficient of thermal expansion = 12\times 10^{-6}\ /^{\circ}C

Change in length is given by

\Delta L=L\alpha \Delta T\\\Rightarrow \Delta L=500\times 12\times 10^{-6}\times (40-(-35))\\\Rightarrow \Delta L=0.45\ m

The change in length is 0.45 m

The number of joints would be

n=\dfrac{0.45}{0.02}\\\Rightarrow n=22.5\\\Rightarrow n\approx 23

The number of joints is 23

Each bridge section length would be

\dfrac{L}{n}=\dfrac{500}{23}=21.7391304348\ m

The length of each bridge section would be 21.7391304348 m

4 0
3 years ago
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